Polarity Protection for PIC board

Hi,

I want to protect a PIC board I have designed and want to polarity protect it. I will have 2 or 3 I/O expansion boards as well and think that I could have upwards of 500mA on the 5V side.

I could try to use a 5A Shottky diode but the leads are so think it will make mounting it a pain.

Is there an easier way?

Thank you,

Ron

ground polarity

What are you trying to protect against?

Your board itself being powered incorrectly, being powered from outside, wrong polarity on output connections or overvoltage?

You can get smaller Shottky diodes but it isn't clear where you intend to connect it.

Brian.

ic 7805 polarity

Hello,

I have a PIC circuit I want to protect against reverse polarity. I have added 4 I2C expander IC's to the 18F4620.

12 - 15VDC input into a LM7805ACT.

I was originally going to put a 5 amp Shottkey, bought some and saw how hard it was to bend the lead to the proper pad spacing. Novices will be making these controllers up so I want it to be fairly easy.

Then I started thinking that a TIP125, base tied to ground through 1K series resistor acts like a switch. After posting here I found mention of MOSFETs being used for this purpose. TIP125 are much less expensive that mosfets.

Would this work?

Thank you,

parallel diodes reverse polarity protection

The 7805 will regulate to 5V with anything more than about 8V at its input so you have plenty of voltage overhead to spare. All you need to do is add a normal silicon diode with a rating of 1 amp or more in line with the input to the regulator.

Diodes like the 1N4001 will work fine. You only need to use Shottky diodes where the voltage drop has to be kept as low as possible, with 4V to spare in your application it isn't worth using one.

If you want to protect your board against reverse polarity coming from somewhere other than your own power supply, again use a silicon diode but this time between the 5V rail and ground with the cathode toward 5V. It will not normally conduct unless the supply polarity is wrong.

Brian.

diode de protection recom 7805

Hello Brian,

I have a 1N4004 in the circuit now. The 1N400x datasheet says that it is rated to 300mW. I could have up to 48 leds (10mA each) connected via the I/O expansion modules so I could draw 500mA at 5VDC.

Will this heat up a 1N400x to much? This is why I thought using a larger Shottky diode would run much cooler.

Then I thought about using a transistor, it seems that it is an ideal solution as the drop through the TIP125 is about 1.4V, as you pointed out 12 - 1.4 = 10.6V which is above the 8VDC required to run the LM7805. It can also handle the current with no problem.


Thank you for the help,

Ron

Added after 29 minutes:

Sorry about the typo....

Thanks for the help Brian..... Fixed the typo in your name......

Ron

using a mosfet as reverse polarity protection

Hello Ron.

I'm not sure where the TIP125 comes into play, using it a a pre-dropper before the 7805 has no advantage whatsoever, all you do is share the heat generated between the transistor and the regulator, the same amount is generated overall. The 7805 is perfectly capable of handling the job on its own.

All you need to worry about is the heat dissipation. Your diode will drop a fairly constant 0.6 volts so you have 11.4V after it. The 7805 drops the 11.4V to 5V so there will be 6.4 volts across it. Assuming worst case, your 500mA load, the diode dissipates 0.6 x 0.5 = 0.3W which is OK and the regulator dissipates 6.4 x 0.5 = 3.2W which is also OK. You will need a heatsink on the regulator though as it will run hot.

If you used your Shottky diode, it would run cooler but the heat would be increased in the 7805 instead. You could consider using a resistor as well as the diode. If we give the 7805 a little extra headroom, say 9V at it's input, 11.4V - 9V could be dumped in a resistor of (2.4 / 0.5) Ohms. You could use a standard 4.7 Ohm resistor with a power rating of (2.4 * 0.5) = 1.2 Watts or greater. The resistor would share the heat if that was beneficial to your design but it might compromise the rise time of your power rail to a small degree.

Don't worry about the typo - I get called all sorts of things

Brian,

how to make pcb reverse polarity protection

Hi Brian,

Where the TIP125 comes in.......

If the TIP125 base is connected to 1K ohm resistor which in turn is connected to ground. Polarity plugged in properly TIP125 conducts and let current through.


If polarity is reversed then the base tied to +12 and TIP125 would not conduct (what about diode in TIP).

Not sure how the diode in the TIP comes into play.

Here is what I found on the web using a mosfet, hoping the same applies for the TIP but I think the voltage drop in TIP is to high.

http://www.geofex.com/Article_Folders/mosswitch/mosswitch.htm

I like the 1N400x idea the best, simple, inexpensive and easy to install.

Thanks,

Ron

reverse polarity protection diode

A couple alternatives to the 1N400x series that offer higher current are the 1N539x (1.5 amp) or the 1N540x (3 amp).

1 ampere diode polarity

Hello,

I want to say thank you to Brian and baysidebecca.

I am going with Brians recommendation and simply using the diode, and I will use the 1.5 amp diode 1N5397 (only .041/100). I will mount it slightly off the PCB in case I draw a little more current than expected (to help with cooling).


Thank you both, your help is greatly appreciated.

Ron

mosfet reverse polarity protection

I will mount it slightly off the PCB in case I draw a little more current than expected (to help with cooling).

Click to expand...

Cooling? The protection diode shall be connected in parallell with the input, not in series with the input.

a series connected diode polarity protection

I disagree VSWR, the diode is in series to block reverse polarity rather than in parallel to shunt reverse current.

halloween_man is tying to prevent damage from a reverse connected supply. In normal use he has 12V available to power a 5V circuit, given that much overhead, putting a diode in series is a viable option. I suggested a parallel diode, reverse biassed, across the 5V rail in case any of his external connections tried to impose reverse voltage, normally of course it wouldn't conduct.

As a series connected diode, it will carry the load current of his design and therefore dissipate (Vf * If) Watts.

Brian.

tip125 heating regulator

You're welcome, Ron! I definetly agree with Brian; the diode should be wired in series with cathode to voltage regulator input. While placing the diode in parallel with the input will protect the circuit for a brief instant, it will also catastrophically destroy the protection diode and possibly some wiring on the circuit board.
One thing I would do is include a fuse in series with the diode as a "final" fail-safe measure. Unless they are soldered into the circuit board, fuses are much easier to replace than a diode and burnt wiring!

baysidebecca

what is polarity protection

Use P-channel enh. MOSFET.

Connect:
Drain ... input Vin+
Gate ... GND
Source ... PIC circuit Vcc+

For more details check attached pdf.

7805 reverse voltage protection

Hai

A simple bridge rectifier will solve the problem and this is the simplest way

Regards
Nandhu

7805 polarity

A bridge will let you connect the polarity either way but it has a major disadvantage over a single diode: the ground potential is raised by the forward voltage drop of one diode. For many applications this is not acceptable, it isn't the voltage loss that's the issue, its the fact that you lose true ground potential.

As for MOSFETs, they are great, they do the job perfectly well ---- and they cost about 20 times as much !!!

Brian.

7805 041 regulator 5v

Hi Brain and Baysidebecca,

I was once told there is a way to polarity protect using a diode and PTC fuse. I have drawn it out but cannot come up with a way that it will work, that I can see anyways.

Any ideas on this?


I am still going with the 1.5 amp diodes in series. These diodes are rated for 4.8 watts. With a 1.4v drop across the diode with a max current draw of 1A is there any reason to be concerned about heat? 1.4v * 1A = 1.4W which is about 33% of the max rating.

I Have ordered my very first prototype boards......... Very excited.....

Thanks

Ron

input reverse voltage protection 5v

Hello Ron.

The PTC is wired in series with the incoming power. A diode (cathode end) is connected at the junction of the PTC and the input of the regulator. The anode end goes to ground. Normally, the diode will not conduct and the PTC thermistor will have low resistance so current flows through it easily.

When a reversed supply is connected, the diode conducts and holds the voltage to around 0.7V, drawing a lot of current. This heats the PTC thermistor and makes its value increase, limiting the current that can flow into the circuit. The PTC thermistor acts like a resetting fuse, it goes high resistance when hot but drops again as it cools down.

Don't be concerned at the power dissipated in a diode. The manufacturers max current rating is all you need to worry about. If it is rated at 1 Amp it implies it can dissipate whatever power it produces at 1 Amp current. Typically the power is around 0.6 to 0.7 multiplied by the current through it.

Brian.